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allsm [11]
3 years ago
6

Helpp!!!!!!!!!!!!!!!​

Mathematics
1 answer:
umka2103 [35]3 years ago
6 0

Answer:

yeah

Step-by-step explanation:

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The vertex of this parabola is at (5, -4). Which of the following could be its equation? A.y = 2(x + 5)2 + 4
astraxan [27]
I think d or b but if I am wrong sorry
4 0
2 years ago
Read 2 more answers
10p - (3p - 4) = 4(p + 1) + 9
andrezito [222]

Answer:

p = 3

Step-by-step explanation:

distribute parenthesis on both sides of the equation

10p - 3p + 4 = 4p + 4 + 9 ( simplify both sides )

7p + 4 = 4p + 13 ( subtract 4p from both sides )

3p + 4 = 13 ( subtract 4 from both sides )

3p = 9 ( divide both sides by 3 )

p = 3


6 0
3 years ago
I have 3 questions, 20 points each please explain your answer and show work, and PLEASE GIVE AN ACTUAL ANSWER. PLEASE ANSWER SOO
alexdok [17]

Answer:

Step-by-step explanation:

1.From the given triangle, ABC, using the proportionality theorem, we get

\frac{FA}{CA}=\frac{EF}{BC}

⇒\frac{8}{40}=\frac{5}{x}

⇒x=25ft

Thus, the height will be 25 feet.

2.  

Statement                                                   Reason

1.∠C≅∠E                                                     Given

2. ∠ABC≅∠DBE                                Vertically opposite angles

3. ΔABC is similar to ΔDBE               AA similarity rule.

5 0
3 years ago
(Number 5)help plzzz
EleoNora [17]

Answer:

See below.

Step-by-step explanation:

So we want to prove that:

\sqrt8+\sqrt2=3\cdot 2^{\frac{1}{2}}

First, simplify √8. This is the same as:

\sqrt8=\sqrt{4\cdot 2}=\sqrt4\cdot\sqrt2=2\sqrt2

Therefore, our equation is now:

2\sqrt2+\sqrt2=3\cdot2^{\frac{1}{2}}

Combine like terms on the left:

3\sqrt2=3\cdot 2^\frac{1}{2}}

The square root of something is the same as taking that number to the one-half power. Thus:

3(2)^\frac{1}{2}}=3\cdot 2^\frac{1}{2}}

Rewrite:

3\cdot2^\frac{1}{2}}\stackrel{\checkmark}{=}3\cdot2^\frac{1}{2}}

And we're done!

6 0
2 years ago
3x^3+x^2-12x-4 by x+2
stepan [7]

Answer:

i think it is c

Step-by-step explanation:

6 0
2 years ago
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